Before we start, let's mention 1 important thing:
Given 4 integers $m_1, n_1, m_2, n_2\in\mathbb{N}$, and 2 matrices $A\in\mathbb{R}^{m_1\times n_1}$, $B\in \mathbb{R}^{m_2\times n_2}$. Assume that(It's not mentioned, but it's obvious, so i assume so.) $A=B$ if and only if $m_1=m_2\wedge n_1=n_2\wedge\forall i\in \{1, \dots, m\}, \forall j\in \{1, \dots, n\}, A_{ij}=B_{ij}.$ We'll use $\text{Definition 1}$ to denote this.

Definition of Elementary Row Operations
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According to the lecture notes given by the Teacher, the definition of 3 types of elementary row operations is
- Type-1. 將矩陣的某兩個 row 對調
- Type-2. 將矩陣的某一個 row 乘上一非零實數
- Type-3. 將矩陣的某一個 row 乘上一實數後加到另一個 row

It's kinda hard for me to make use of it. Despite the doubt about the reason why multiplying a "row" by a scalar is valid, as vector space and row vector is not mentioned yet, I still find difficulty writing the proof.

I'll take the following for example.

Example: Basic Property of Elementary Row Operations.
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Given 3 positive integers $k$, $m$, $n$, and 3 functions $f_1$, $f_2$ and $g$ from $\mathbb{R}^{m\times n}$ to $\mathbb{R}^{m\times n}$ that performs elementary row operations:
- $f_1$ performs $type\ 2$ elementary row operation on a matrix, multiplying $k$-th row of which by $3$.
- $f_2$ performs $type\ 2$ elementary row operation on a matrix, multiplying $k$-th row of which by $2$.
- $g$ performs $type\ 2$ elementary row operation on a matrix, multiplying $k$-th row of which by $6$.

Show that $f_2 \circ f_1 = g$ i.e. $\forall A\in\mathbb{R}^{m\times n}, (f_2 \circ f_1)(A)=g(A)$.

Remember, to prove the equality holds for any matrix with same size, we should apply $\text{Definition 1}$ to the statement, and then make uses of the definition of elementary row operations.

Here's my answer to this question: Assume that performing $type\ 2$ elementary row operation makes each entries on the specific row multiplied by a non-zero number.
Given $A\in \mathbb{R}^{m\times n}$, Let $B_1 = f_1(A)$, $B_2 = f_2(B_1)$, $C = g(A)$.
Then according to the assumption, we can now describe the entries of $B_1$: $${B_1}_{ij} = \begin{cases}A_{ij} & \text{if} & i\not=k \\ 3A_{ij} & \text{if} & i=k \end{cases}.$$ We can also describe the rest: $$\begin{aligned}{B_2}_{ij} &= \begin{cases}{B_1}_{ij} & \text{if} & i\not=k \\ 2{B_1}_{ij} & \text{if} & i=k \end{cases} \\ &= \begin{cases}A_{ij} & \text{if} & i\not=k \\ 6A_{ij} & \text{if} & i=k \end{cases}\end{aligned}$$ and $$C_{ij} = \begin{cases}A_{ij} & \text{if} & i\not=k \\ 6A_{ij} & \text{if} & i=k \end{cases}.$$ Thus it's obvious that $\forall i\in \{1, \dots, m\}, \forall j\in \{1, \dots, n\}, A_{ij}=B_{ij}$, and thus by $\text{Definition 1}$, $B_2 = C$, and thus $f_1 \circ f_2 = g.\blacksquare$

Reflection
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When solving a system of linear equations it won't cause any inconvenience. When it comes to proving a basic property of elementary row operations, it seems like we can still handle it, but what if there's more?

What is the proper way of making use of the definition of the elementary row operations? What do you think?